在Python中获取矩阵/列表中的所有对角线

Pythonic方法

对于纯Python实现，我建议使用1D方式进行操作。

W, H = len(mat[0]), len(mat) 
idx = range(W-1) + range(W-1, W*H, W)
rng = range(1, W) + range(H, 0, -1)
rng = map(lambda x: x if (x < min(W, H)) else min(W, H), rng)
dia = [[i + (W-1) * m for m in xrange(r)] for i, r in zip(idx, rng)]

这里dia返回每条对角线的索引列表。要检索相应的值：

arr = [e for row in mat for e in row] #Flatten the matrix
for d in dia:
    print [arr[e] for e in d][::-1]

[-2]
[9, 5]
[3, -6, 3]
[-1, 2, 5, 2]
[8, 7, 1]
[-4, 3]
[8]

arr2 = [e for row in zip(*mat[::-1]) for e in row] #Flatten and rotate the matrix by 90°
for d in dia[::-1]:
    print [arr2[e] for e in d]

[2]
[3, 1]
[5, 5, 3]
[-2, -6, 7, 8]
[9, 2, -4]
[3, 8]
[-1]

tril = [np.flip(np.fliplr(mat).diagonal(n)) for n in xrange(mat.shape[0])][::-1]
trir = [np.flipud(mat).diagonal(n) for n in xrange(1, mat.shape[0])]
dia = tril + trir

[array([-2]),
 array([9, 5]),
 array([ 3, -6,  3]),
 array([-1,  2,  5,  2]),
 array([8, 7, 1]),
 array([-4,  3]),
 array([8])]

Try this :

import numpy as np
matrix = [[-2,  5,  3,  2],
          [ 9, -6,  5,  1],
          [ 3,  2,  7,  3],
          [-1,  8, -4,  8]]

matrix = np.array(matrix)
matrix = np.flipud(matrix)
a = matrix.shape[0]
list_ = [np.diag(matrix, k=i).tolist() for i in range(-a+1,a)]
print(list_)

输出：

[[-2], [9, 5], [3, -6, 3], [-1, 2, 5, 2], [8, 7, 1], [-4, 3], [8]]

使用一些numpy-fu来获取主对角线：

import numpy as np  
r = np.arange(36) 
r.resize((6, 6)) 
print(r) 
r = r.reshape(len(r)**2)[::len(r)+1] 
print(r)

输出：

[[ 0  1  2  3  4  5]
 [ 6  7  8  9 10 11]
 [12 13 14 15 16 17]
 [18 19 20 21 22 23]
 [24 25 26 27 28 29]
 [30 31 32 33 34 35]]

[ 0  7 14 21 28 35]

从这里开始：np.Diagonal

 np.diagonal(matrix)